
A
minimum of algae is when we derive and equal 0
Derivation

equal 0

the amout of algae reach minimum when t=36/7
B
to calculate the minumum number of algae we replace the days when algae reach minumum(previus exercise) on function

the rounded minimum number of algae is 124
C
We replace the amount of algae to 900 on the original function

simplify

now solve t factoring by
![x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://img.qammunity.org/2023/formulas/mathematics/college/rxvf73usjbbwyik14knxdemoz21vfz2ufc.png)
where x is the variable then t, a is 35, b is -360 and c 150
replacing
![t=\frac{-(-360)\pm\sqrt[]{(-360)^2-4(35)(150)}}{2(35)}](https://img.qammunity.org/2023/formulas/mathematics/college/54wgxto3q9v90sa1ongd3p5h4n96tlvpme.png)
simplify
![\begin{gathered} t=\frac{360\pm\sqrt[]{129600-21000}}{70} \\ \\ t=\frac{360\pm\sqrt[]{108600}}{70} \\ \\ t=\frac{360\pm\sqrt[]{100*1086}}{70} \\ \\ t=\frac{360\pm10\sqrt[]{1086}}{70} \\ \\ t=\frac{36\pm\sqrt[]{1086}}{7} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jsqnra5x7newbx461wmzjdpmvdsfvs93fs.png)
then the days to have 900 algae is
![\begin{gathered} t_1=\frac{36+\sqrt[]{1086}}{7}\approx9.85 \\ \\ t_2=\frac{36-\sqrt[]{1086}}{7}\approx0.46 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3xc86tv0ybooqv66ejrbd5t6m8oz818wb6.png)
we have 2 values for 900 algae 9.85 and 0.46 days
D Graph
we know the graph is a parable because maximum exponent is 2
we know points like,
minimum amount of algae A(5.14 , 124.3) ,
when the amount of algae is 900 B(9.85 , 900) C(0.46 , 900)
then we can palce the points of a graph and join them